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compact subspaces in Hausdorff spaces are separated by neighbourhoods from points

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Topology

topology (point-set topology, point-free topology)

see also differential topology, algebraic topology, functional analysis and topological homotopy theory

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Statement

Lemma

(compact subspaces in Hausdorff spaces are separated by neighbourhoods from points)

Let

  1. (X,τ)(X,\tau) be a Hausdorff topological space;

  2. YXY \subset X a compact subspace.

Then for every xX\Yx \in X \backslash Y there exists

  1. an open neighbourhood U x{x}U_x \supset \{x\};

  2. an open neighbourhood U YYU_Y \supset Y

such that

  • they are still disjoint: U xU Y=U_x \cap U_Y = \emptyset.

As a direct consequence compact subspaces of Hausdorff spaces are closed.

Proof

By the assumption that (X,τ)(X,\tau) is Hausdorff, we find for every point yYy \in Y disjoint open neighbourhoods U x,y{x}U_{x,y} \supset \{x\} and U y{y}U_y \supset \{y\}. By the nature of the subspace topology of YY, the restriction of all the U yU_y to YY is an open cover of YY:

{(U yY)Y} yY. \left\{ (U_y \cap Y) \subset Y \right\}_{y \in Y} \,.

Now by the assumption that YY is compact, there exists a finite subcover, hence a finite set SYS \subset Y such that

{(U yY)Y} ySY \left\{ (U_y \cap Y) \subset Y \right\}_{y \in S \subset Y}

is still a cover.

But the finite intersection

U xsSYU x,s U_x \coloneqq \underset{s \in S \subset Y}{\cap} U_{x,s}

of the corresponding open neighbourhoods of xx is still open, and by construction it is disjoint from all the U sU_{s}, hence also from their union

U YsSYU s. U_Y \coloneqq \underset{s \in S \subset Y}{\cup} U_s \,.

Therefore U xU_x and U YU_Y are two open subsets as required.

Created on May 19, 2017 at 10:10:29. See the history of this page for a list of all contributions to it.